Let $f(x)=x^3-6x^2+12x$ and let $c$ be the number that satisfies the Mean Value Theorem for $f$ on the interval $[0,3]$. What is $c$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $1$ (Choice C) C $2$ (Choice D) D $3$
Solution: According to the Mean Value Theorem, there exists a number $c$ in the open interval $(0,3)$ such that $f'(c)$ is equal to the average rate of change of $f$ over the interval: $f'(c)=\dfrac{f(3)-f(0)}{(3)-(0)}$ First, let's find that average rate of change: $\dfrac{f(3)-f(0)}{(3)-(0)}=\dfrac{9-0}{3}={3}$ Now, let's differentiate $f$ and find the $x$ -value for which $f'(x)={3}$. $f'(x)=3x^2-12x+12$ The solutions of $f'(x)=3$ are $x=1$ and $x=3$. Out of these, only $x=1$ is within the interval $(0,3)$. In conclusion, $c=1$.